Answer:
y - 1 = 0
Step-by-step explanation:
move constant to the left by adding its opposite to both sides y - 1 = 1 - 1
the sum two opposites equals 0
y =1
Answer:
The answer is x=10
Step-by-step explanation:
5/6x - 10 = 1/3x -5 multiple both sides by 6
5x - 60 = 2x -30 move the terms
5x - 2x = -30 +60
3x = 30 divide by 3
x = 10
Answer:
-127.488°
Step-by-step explanation:
45" is 0.75 of 1', so -128° 30' 45" = -128° 30.75'
Next: 30.75' is 30.75/60 of 1°, and or -128° plus 0.5125 of 1°.
The final answer is found by adding 0.5125° to -128°:
-128.0000°
+ 0.5125°
--------------------
-127.4875
To the nearest thousandths of a degree, that would be -127.488°
Area = π x radius²
dA/dr = 2πr
Average rate of change = (rate of change at one instant + rate of change at other instant)/2
= (2πr₁ + 2πr₂)/2
= (r₁ + r₂)π
Ai) 9π
Aii) 8.5π
Aiii) 8.1π
B) Instantaneous change:
(dA/dr) at (r = 4) = 2π(4)
= 8π
Answer:
r=294.9m
Step-by-step explanation:
The forces on the particle are
Now , we sum all these forces to get the net force
we can use the fact F=m*a and integrate the acceleration
![a(t)=\frac{1}{m}F(t)\\\\v(t)=\int a(t)dt=\frac{1}{m}\int{F_{T}}dt\\\\v(t)=\frac{1}{m}[(57t-t^{2}+\frac{5}{3}t^{3})\hat{i}+(12t-2t^{2})\hat{j}+(-t^{2}-t)\hat{k}]\\\\r(t)=\int v(t)dt=\frac{1}{m}[(\frac{57}{2}t^{2}-\frac{1}{3}t^{3}}+\frac{5}{4}t^{4})\hat{i}+(6t^{2}-\frac{2}{3}t^{3})\hat{j}+(-\frac{1}{3}t^{3}-\frac{1}{2}t^{2})]](https://tex.z-dn.net/?f=a%28t%29%3D%5Cfrac%7B1%7D%7Bm%7DF%28t%29%5C%5C%5C%5Cv%28t%29%3D%5Cint%20a%28t%29dt%3D%5Cfrac%7B1%7D%7Bm%7D%5Cint%7BF_%7BT%7D%7Ddt%5C%5C%5C%5Cv%28t%29%3D%5Cfrac%7B1%7D%7Bm%7D%5B%2857t-t%5E%7B2%7D%2B%5Cfrac%7B5%7D%7B3%7Dt%5E%7B3%7D%29%5Chat%7Bi%7D%2B%2812t-2t%5E%7B2%7D%29%5Chat%7Bj%7D%2B%28-t%5E%7B2%7D-t%29%5Chat%7Bk%7D%5D%5C%5C%5C%5Cr%28t%29%3D%5Cint%20v%28t%29dt%3D%5Cfrac%7B1%7D%7Bm%7D%5B%28%5Cfrac%7B57%7D%7B2%7Dt%5E%7B2%7D-%5Cfrac%7B1%7D%7B3%7Dt%5E%7B3%7D%7D%2B%5Cfrac%7B5%7D%7B4%7Dt%5E%7B4%7D%29%5Chat%7Bi%7D%2B%286t%5E%7B2%7D-%5Cfrac%7B2%7D%7B3%7Dt%5E%7B3%7D%29%5Chat%7Bj%7D%2B%28-%5Cfrac%7B1%7D%7B3%7Dt%5E%7B3%7D-%5Cfrac%7B1%7D%7B2%7Dt%5E%7B2%7D%29%5D)
and we evaluate in r(2) an we take the norm to obtain the distance
![r(2)=\frac{1}{m}[\frac{394}{3}\hat{i}+\frac{56}{3}\hat{j}-\frac{14}{3}\hat{k}]\\|r(2)|=\frac{1}{m}\sqrt{[(\frac{394}{3})^{2}+(\frac{56}{3})^{2}+(\frac{14}{3})^{2}]}\\|r(2)|=\frac{132.73}{0.45}=294.9m](https://tex.z-dn.net/?f=r%282%29%3D%5Cfrac%7B1%7D%7Bm%7D%5B%5Cfrac%7B394%7D%7B3%7D%5Chat%7Bi%7D%2B%5Cfrac%7B56%7D%7B3%7D%5Chat%7Bj%7D-%5Cfrac%7B14%7D%7B3%7D%5Chat%7Bk%7D%5D%5C%5C%7Cr%282%29%7C%3D%5Cfrac%7B1%7D%7Bm%7D%5Csqrt%7B%5B%28%5Cfrac%7B394%7D%7B3%7D%29%5E%7B2%7D%2B%28%5Cfrac%7B56%7D%7B3%7D%29%5E%7B2%7D%2B%28%5Cfrac%7B14%7D%7B3%7D%29%5E%7B2%7D%5D%7D%5C%5C%7Cr%282%29%7C%3D%5Cfrac%7B132.73%7D%7B0.45%7D%3D294.9m)
I hope this is useful for you
regards
